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Some reference in adaptive techniques say that when the solution $u$ is smooth enough we can use p-refinement instead of h-refinement. And when we have for example singularity, we should use h-refinement. I can not understand it. why? When we use a higher order polynomial for approximation we can get smoother solution, so it seems there is no problem for using p-refinement.

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  • $\begingroup$ Are you asking for a mathematical justification or some kind of intuitive idea? $\endgroup$ – knl Sep 17 '17 at 18:41
  • $\begingroup$ No, I just want to know the mathematical justification. $\endgroup$ – Rosa Sep 17 '17 at 18:43
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The decision whether to do p- or h- adaptivity is to achieve faster, potentially exponential convergence rate. In other words to get a solution with given error with minimal computational effort.

Limiting considerations to elliptic problems. The approximate solution can converge to exact solution if you do h- or p- adaptivity. If the exact solution is smooth, convergence is faster if you do p- adaptivity. To see this look at a priory error estimator, \begin{equation} \| u - u^h \| \leq C h^p \left\| \frac{\partial^{p+1} u}{\partial x^{p+1}} \right\| \end{equation} where $p$ is polynomial order of base functions. If the solution is smooth, higher order derivatives fast approach zero, so that a priory error estimator. So if you increasing order of base functions, you very quickly converge to an exact solution.

Unfortunately, that is not working if the solution is not regular, for example if you have a singularity. Higher order derivatives are not bounded, and p- adaptivity give a solution with larger computational cost comparing to h- adaptivity.

As an example see L-shape body with transport problem, enter image description here enter image description here For details see here http://mofem.eng.gla.ac.uk/mofem/html/mix_transport.html. Mix formulation is applied there, but the same goes for classical finite elements. You can see that both h- and p- adaptivity show convergence, however the fastest convergence you get if you locally at singularity do h- refinement.

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  • $\begingroup$ Great answer. But in the formula, the exponent $\mu$ should be $p$, and you need the $(p+1)$th derivative instead of the $p$th. $\endgroup$ – Wolfgang Bangerth Sep 17 '17 at 22:43
  • $\begingroup$ @WolfgangBangerth You right, thanks for spotting this. If the exact solution is polynomial of order p and approximation base functions is order p, a priori error estimator is 0. And it is if (p+1)th derivative is there. $\endgroup$ – likask Sep 17 '17 at 22:51

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